[Aritra]

if you tile the plane by congruent regular polygons, there must be nn polygons meeting at each vertex. Thus the interior angles of each polygon must be 2π/n, for some positive integer n .

now see that for n≤6 the angle is greater than π/3 .

but for , n>6 the polygons would need to have angles less than π/3, which is impossible.

but when n= 5 then internal angle is 108 which doesnot divide 360 so pentagon is not possible   ,  option a is correct

[Aritra]

i think [.] denotes the greatest integer function write , other wise the problem has no meaning . now see that a floor function is dis continuous in all those integer value

([6sin(x)]) can take integer value only when (sin (x)) take the value multiple of (\frac{n}{6}) see that n= 0,1,2,3,4,5,6

now (\frac {1}{6}) can be obtained by sin x only once at the point x= (\frac{\pi}{2})

and all the value are attended twice once at 0 to\ (\frac{pi}{2}) and another  (\frac{\pi}{2}) to (\pi)  so total there are 12 point of discontinuity

i don't think it is continuous at x=0 because at x=0 it takes a integer value and floor function is not continuous at any integer value

even the graph say that it has 12 point of discontinuity

https://www.desmos.com/calculator/wnacsak92c

please clarify if i am wrong at any point

 

[Aritra]

this is your required diagram and the proof is quite trivial

[Aritra]

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